16- Write a program to swap the first number with last number in an integer array elements A

16- Write a program to swap the first number with last number in an integer array elements A

Hello I'm from a non CS/IT field, interested to learn coding and looking forward to build a career in the IT field. I've started with c language. Since I'm an absolute beginner, I'd highly appreciate any suggestions/advice on how I should approach learning and practicing c language, as well as improving my cv/resume (and other requirements in any, like some programming languages or tools that I need to learn) so as to become employable in the IT sector. Thank you

You can even try coding through sololearn and freecodeacademy for free but it would be good if anyone recommend here you a good videos (channel) or (e)books to learn as I am not familiar with it. But start with C then you can learn other language easily as C is the mother of all the languages

You can even try coding through sololearn and freecodeacademy for free but it would be good if anyone recommend here you a good videos (channel) or (e)books to learn as I am not familiar with it. But start with C then you can learn other language easily as C is the mother of all the languages

#include<iostream> using namespace std; int main() { int y,m; cout <<"please enter year"<<endl; cin>>y; cout <<"please enter month"<<endl; cin>>m; switch(m) { case 2:(y%4==0 && y&100 !=0 || y%400==0); cout <<"the month has 29 days"<<endl; break; case 4: case 6: case 9: case 11: cout <<"the month has 30 days"<<endl; break; case 1: case 5: case 3: case 7: case 8: case 12: cout <<"the month has 31 days "<<endl; break; default:cout<<"not valid"; } return 0; }

Hello , could someone please suggest me some resources to solve rectangle intersection problem , video or book or maybe some forum , thanks in advance

#include<iostream> using namespace std; int main() { int y,m; cout <<"please enter year"<<endl; cin>>y; cout <<"please enter month"<<endl; cin>>m; switch(m) { case 2:(y%4==0 && y&100 !=0 || y%400==0); cout <<"the month has 29 days"<<endl; break; case 4: case 6: case 9: case 11: cout <<"the month has 30 days"<<endl; break; case 1: case 5: case 3: case 7: case 8: case 12: cout <<"the month has 31 days "<<endl; break; default:cout<<"not valid"; } return 0; }

but ı want to use with pin interrupt this button

On which Microcontroller?

#include<iostream> using namespace std; int main() { int y,m; cout <<"please enter year"<<endl; cin>>y; cout <<"please enter month"<<endl; cin>>m; switch(m) { case 2:(y%4==0 && y&100 !=0 || y%400==0); cout <<"the month has 29 days"<<endl; break; case 4: case 6: case 9: case 11: cout <<"the month has 30 days"<<endl; break; case 1: case 5: case 3: case 7: case 8: case 12: cout <<"the month has 31 days "<<endl; break; default:cout<<"not valid"; } return 0; }

nuvoton n76e003

case 2: if(!(y%400) || (!(y%4)&&(y%100))) cout << "29 days"; else cout << "28 days"; break;

#include <iostream> #include <chrono> #include <format> int main() { std::chrono::year_month y_m; std::cout << "Enter yyyy mm: "; std::chrono::from_stream(std::cin, "%Y %m", y_m); if (y_m.ok()) { std::cout << std::format("The month has {} days.\n", (y_m / std::chrono::last).day()); } else { std::cout << "Invalid month!\n"; } }

The idea is to help people with their code with maybe suggestions on how to improve it. Not to go off in a tangential direction and give them alternative code which may or may not even work on their compilers 🙂. Btw, most of the major organizations are yet to adapt C++17 itself. C++20 is a long way off especially for students because most schools still wont be using the latest compilers.

16- Write a program in c++ to swap the first number with last number in an integer array elements A

ye but people often don't even know that these parts of C++ even exist. most schools would force Turbo C++ with no standard library other than iostream related facilities if they could.

16- Write a program in c++ to swap the first number with last number in an integer array elements A

16- Write a program in c++ to swap the first number with last number in an integer array elements A

The students dont have a choice. You can help them with their current code and point them to materials where they can learn about modern C++.

https://helloacm.com/rectangle-intersection-test/

set(Num 2) set(two 2) set(tree 3) string(REPLACE ${two} ${tree} res Num) message("result is ${${res}}") I want to replace text but cmake doesn't work am i do it right ?

You should be using ${Num}

you mean string(REPLACE ${two} ${tree} res ${Num}) right ?

you mean string(REPLACE ${two} ${tree} res ${Num}) right ?

got it i should change this line message("result is ${${res}}") to this message("result is ${res}")

ye but people often don't even know that these parts of C++ even exist. most schools would force Turbo C++ with no standard library other than iostream related facilities if they could.

I didnt even notice that line. 😁

time complexity O(n) ?

The idea is to help people with their code with maybe suggestions on how to improve it. Not to go off in a tangential direction and give them alternative code which may or may not even work on their compilers 🙂. Btw, most of the major organizations are yet to adapt C++17 itself. C++20 is a long way off especially for students because most schools still wont be using the latest compilers.

for( i=2; count<=num; i++ ) { int n = i; while (n % 2 == 0) n = n/2; while (n % 3 == 0) n = n/3; while (n % 5 == 0) n = n/5; if( n == 1 ) { count++; if(count == num) return i; } }

The time complexity is not directly proportional to num. It depends on how numbers that have only 2,3 and 5 as prime factors are distributed.

for( i=2; count<=num; i++ ) { int n = i; while (n % 2 == 0) n = n/2; while (n % 3 == 0) n = n/3; while (n % 5 == 0) n = n/5; if( n == 1 ) { count++; if(count == num) return i; } }

The time complexity is not directly proportional to num. It depends on how numbers that have only 2,3 and 5 as prime factors are distributed.

time complexity O(n) ?

I guess its O(n * 2^n) (not tight bound). And Ω(n lg n). Here n is num.

O(n)

It doesn’t matter; complexity is not associative.

Time complexity in this case is not dependent on the input. So something like n or n log n wont work here. The time complexity depends on the distribution. For ex running that algorithm for num = 100 in the range say 1 to 10000 would be a lot faster than running it in between say 2 consecutive zeroes of riemann zeta function say after the 1000th bernoulli number

O(n) * [ O(log2 n) + O(log3 n) + O(log5 n) ] = n log(n) in coding we treat log as log of base 2

O(n) * [ O(log2 n) + O(log3 n) + O(log5 n) ] = n log(n) in coding we treat log as log of base 2

Time complexity in this case is not dependent on the input. So something like n or n log n wont work here. The time complexity depends on the distribution. For ex running that algorithm for num = 100 in the range say 1 to 10000 would be a lot faster than running it in between say 2 consecutive zeroes of riemann zeta function say after the 1000th bernoulli number

Time complexity in this case is not dependent on the input. So something like n or n log n wont work here. The time complexity depends on the distribution. For ex running that algorithm for num = 100 in the range say 1 to 10000 would be a lot faster than running it in between say 2 consecutive zeroes of riemann zeta function say after the 1000th bernoulli number

But its not random distribution. This complexity requires some math to get it. But its not impossible. (I assume the input is 'num').

That is why I said it is not just a simple direct proportionality relation to any of the complexities that we regularly use like n, nlgn, n^2 and so on. It depends on the distribution of such numbers. And the math is not that complicated. I can show it here but unfortunately this is not the group for such discussions.

If the idea was to just find out the nth number that just has 2,3 and 5 as prime factors, then there are much better and simpler algorithms out there which are linear. The algorithm provided in the OP's question is not however efficient and determining its complexity is slightly tougher.

That is why I said it is not just a simple direct proportionality relation to any of the complexities that we regularly use like n, nlgn, n^2 and so on. It depends on the distribution of such numbers. And the math is not that complicated. I can show it here but unfortunately this is not the group for such discussions.

Can anyone explain this how , n^logn= O(2^n)

hi guys, can someone help me? I created a makefile with the Visual Studio Code editor. The makefile is called "Makefile" and is a MAK file. The problem is that with the make command from the terminal it does not find any makefiles. Yet I am in the folder where it is saved ... What should I do ?

plz send the efficient algorithm

O(n) * [ O(log2 n) + O(log3 n) + O(log5 n) ] = n log(n) in coding we treat log as log of base 2